Given that \( x + y = 100 \), we aim to find the largest possible \(\gcd(x, y)\). Let \( d = \gcd(x, y) \). Then \( x = dm \) and \( y = dn \) for some integers \( m \) and \( n \) such that \(\gcd(m, n) = 1\). Substituting into the sum, we have:

["# Maximizing (\gcd(x, y)) When ( x + y = 100 )", "Finding the largest possible greatest common divisor (gcd) of two positive integers ( x ) and ( y ) whose sum is fixed—here ( x + y = 100 )—is a classic number theory problem with elegant solutions rooted in divisibility and prime factors.", "Given that ( x + y = 100 ), we seek the maximum value of ( d = \gcd(x, y) ). Since ( d ) divides both ( x ) and ( y ), it must divide their sum:", "[\nd \mid (x + y) = 100\n]", "So ( d ) is a positive divisor of 100. The largest possible ( \gcd(x, y) ) cannot exceed 100, and since both ( x ) and ( y ) are less than 100 (unless one is 100 and the other is 0—which is excluded since we consider positive integers), the maximum feasible ( d ) is at most 50.", "Let ( d = \gcd(x, y) ). Then we can write:", "[\nx = d \cdot m, \quad y = d \cdot n\n]", "where ( m ) and ( n ) are positive integers with (\gcd(m, n) = 1) (since the gcd of (x) and (y) is exactly (d)). Substituting into the sum:", "[\nx + y = d(m + n) = 100\n]", "Thus,", "[\nm + n = \frac{100}{d}\n]", "Our goal is to maximize ( d ) such that ( \frac{100}{d} ) is an integer and there exist coprime positive integers ( m ) and ( n ) summing to ( \frac{100}{d} ).", "---", "### Key Insight: Existence of Coprime Pairs Summing to ( k )", "For any integer ( k \geq 2 ), there exist coprime positive integers ( m ) and ( n ) such that ( m + n = k ) if and only if ( k ) is not a power of 2.", "Why?\n- For odd ( k ), pick ( m = 1 ), ( n = k - 1 ). Since ( k ) is odd, ( \gcd(1, k-1) = 1 ), so they are coprime.\n- For even ( k ): if ( k ) is a power of 2 (e.g., 2, 4, 8, 16, 32, 64), we cannot use ( m=1, n=k-1 ) unless ( k-1 = 1 ), which fails because ( k-1 = 3 ) when ( k = 4 ), and ( \gcd(1,3)=1 ), but for ( k = 8 ), ( m=1, n=7 ) works. However, deeper analysis shows that while some even powers of 2 allow coprime pairs, more importantly, for ( k ) a power of 2 greater than 2, we must verify whether all such ( k ) support a coprime pair. Actually, most even ( k > 2 ) do allow a coprime pair: just pick ( m = 1 ), ( n = k - 1 ), and since ( k ) even, ( \gcd(1, k-1) = 1 ). Therefore, the only value of ( k ) for which no coprime pair ( m+n = k ) exists is when ( k = 1 ) or ( k = 2 ) — but ( m,n \geq 1 ) implies minimum sum is 2.", "Wait: if ( k = 2 ), ( m = 1, n = 1 ), and ( \gcd(1,1) = 1 ), so it’s valid.\nIf ( k = 1 ), invalid since ( m,n \geq 1 ).\nSo for all ( k \geq 2 ), there does exist at least one pair ( m,n ) with ( m+n = k ) and ( \gcd(m,n)=1 ). In fact, ( m=1, n=k-1 ) always works because ( \gcd(1, k-1) = 1 ).", "But here’s the catch: ( x = dm \geq d \geq 1 ), ( y = dn \geq d ), and since ( x + y = 100 ), we must ensure ( dm < 100 ), ( dn < 100 ). However, since ( m + n = 100/d ), and ( m, n \geq 1 ), their values are small—maximum possible ( m ) or ( n ) is ( 99/d ), so ( x = d \cdot m \leq d \cdot (99/d) = 99 < 100 ). Therefore, both ( x ) and ( y ) are strictly less than 100, so positivity and integrality are satisfied as long as ( m, n \geq 1 ).", "Thus, the only constraint is that ( d \mid 100 ), and ( m + n = 100/d ) must be expressible as sum of two coprime positive integers—which as shown, is always possible for every integer ( k \geq 2 ), including ( k = 50 ).", "Hence, the maximum ( d ) is the largest divisor of 100 such that ( \frac{100}{d} \geq 2 ), which is always true for ( d \leq 50 ).", "The largest divisor of 100 less than or equal to 50 is 50.", "Try ( d = 50 ):", "Then ( m + n = 100 / 50 = 2 ). Only possibility: ( m = 1, n = 1 ).\nThen ( \gcd(m, n) = \gcd(1,1) = 1 ), valid.\nSo ( x = 50 \cdot 1 = 50 ), ( y = 50 \cdot 1 = 50 ), and ( \gcd(50, 50) = 50 ).", "Thus, ( d = 50 ) is achievable.", "Is a larger ( d ) possible? The next divisor would be 100, but ( d = 100 \Rightarrow m + n = 1 ), impossible with positive integers. So 50 is the maximum.", "---", "### Conclusion", "The largest possible ( \gcd(x, y) ) for positive integers ( x, y ) with ( x + y = 100 ) is 50, achieved when ( x = y = 50 ). This result stems from expressing ( x = dm ), ( y = dn ), requiring ( d \mid 100 ), and leveraging that for every divisor ( d ), a coprime pair ( (m,n) = (1, k-1) ) exists with ( k = 100/d \geq 2 ).", "Final Answer:\n[\n\boxed{50}\n]", "---", "Keywords for SEO:\n( \gcd(x, y) ), maximize (\gcd(x, y)), ( x + y = 100 ), largest gcd, number theory, divisors of 100, coprime integers, integers (x, y), sum constraint, (\gcd) maximization."]